gl & hf

another hope not back to green :(

Why does my 6839434 for A fail?

This round is not second. This round is the ninth my round

You rating curve is so interesting~

http://mirror.codeforces.com/blog/gridnevvvit

start with C. This worked with me, hope it works with you also.

I think they are people which already have an old account at Codeforces but they just build a new account to become candidate master in 1 contest!

let them
I think making another account to compete in a div2 contest and be from top 10 ,that's the result of competing at div1 and lose every time or you can't make a significant score so you leave the real competing at div1 and try to stretch your muscles on a div2 players :)

Black? Dat's racist.

http://mirror.codeforces.com/blog/entry/12378#comment-170490

NO.

so mainstream ;)

Sure. ^_^

I think that's a bad idea according to amount of trolls around :D

Согласен, но зачем нужен гугл? Я нашел четкую формулу, чтобы высчитать ожидание

What thing is unclear for you in the statements? I will try to do better statements in the future

Did you find the exit you asked for?

wasn't unclear ... just a little bit hard to understand . ;)

yaap... statement was not clear to me too... :/
but the notes after each problems helps me a lot... thanks to author for that... :)

There were a few key points :
1) Number of swaps needed to make a permutation an identity permutation are N-C, where N is the number of elements and C is the number of cycles in the permutation. This can be proved using the statement 2 and 3 below (that it is unoptimal to swap outside cycle) and that a cycles of size M needs M-1 swaps to be sorted (provable by induction on M).

2) Swapping two numbers in the same permutation cycle increases the number of cycles by 1. This can be proved constructively (take a cycle C, swap two elements and see that two new cycles are made.)

3) Swapping two number from different permutations decreases the number of cycles by 1. Again, constructive proof.

We have some number of cycles in the given permutation P, and we need N-M cycles in the target permutation Q. Hence keep on making an inter or intra cycle swap depending on which is bigger; and greedily choose the smallest index for lexicographically smallest solution. Complexity O(N²)

you must sort your trees by days and then model situation

so what is the good test to hack them?

Till max day + 1 of course :))

Good hack.. But in my room almost everybody run till <=3001 :) I have hacked only 3 submission.

Hope that the rating system's to!

congrats for that

I'm old to CF and I also read problem E first...

at first you should sort them by day.

have three integer variables curPicked, prevDay and curDay. Then try to pick as many fruits as possible from the trees of prevDay and curDay while making sure that curPicked<=v.

In my solution I created an array of 3000 elements for the days and for each day I saved all the different amounts of fruits that can be gathered on that day specifically. Note that the order doesn't matter.

then it's just a matter of using the above 3 variables and the array effectively while taking care of the corner cases.

Total time complexity: Θ(max of n)

I see no reason, why you assumed that only for any particular day x, only one tree will have fruit at that day. Given your piece of code, day[st[i].day]=st[i].cap;.

corrected...the days were repeated...did'nt noticed it just changed day[st[i].day]=st[i].cap; to day[st[i].day]+=st[i].cap; and AC :'(

me too

yes

I guess in the first code, you can still collect fruits that became available on day 3000, on day 3001, which you never visit, the second code, allowed for this case to occur, by having +2 beyond the maximum day.

Edited code of your first one : 6849997
just written 3001 instead of 3000. Hope, you have got the point.

I tried solving the problem greedily

we need k tubes, each with size >=2. Now k-1 tubes will be of size 2 and the remaining points will be the last tube. However I didn't have enough time to figure out how to print the final tube in a way that satisfies the condition

for any integer i (1 ≤ i ≤ r - 1) the following equation |xi - xi + 1| + |yi - yi + 1| = 1 holds;

Got it!, this is not optimal, because I should go for the trees that are getting rotten first.